A body moving with a velocity of 15 m/s is uniformly accelerated at 3 m/s2. Calculate its velocity after travelling 20 m.

Here are the solutions to the problems: 1. A body moving with a velocity of 15 m/s is uniformly accelerated at 3 m/s^2. Calculate its velocity after travelling 20 m. Given: Initial velocity, u = 15 m/s Acceleration, a = 3 m/s^2 Displacement, s = 20 m Final velocity, v = ? Step 1: Use the kinematic equation v^2 = u^2 + 2as. v^2 = (15 m/s)^2 + 2(3 m/s^2)(20 m) v^2 = 225 m^2/s^2 + 120 m^2/s^2 v^2 = 345 m^2/s^2 Step 2: Solve for v. v = sqrt(345 m)^2/s^2 v ≈ 18.57 m/s The final velocity is 18.57 m/s. --- 2. A particle starts from rest and covers 45 m in 5 sec under uniform acceleration. Determine the acceleration. Given: Initial velocity, u = 0 m/s (starts from rest) Displacement, s = 45 m Time, t = 5 s Acceleration, a = ? Step 1: Use the kinematic equation s = ut + (1)/(2)at^2. 45 m = (0 m/s)(5 s) + (1)/(2)a(5 s)^2 45 m = 0 + (1)/(2)a(25 s^2) 45 m = 12.5a s^2 Step 2: Solve for a. a = 45 m12.5 s^2 a = 3.6 m/s^2 The acceleration is 3.6 m/s^2. --- 3. A train travelling at 72 km/h is brought to rest in 10 seconds. Determine: (i) the acceleration (m/s^2) (ii) the distance travelled before stopping (m) Given: Initial velocity, u = 72 km/h Final velocity, v = 0 m/s (brought to rest) Time, t = 10 s Step 1: Convert initial velocity from km/h to m/s. u = 72 km/h × 1000 m1 km × 1 h3600 s u = 72 × (1000)/(3600) m/s u = 20 m/s (i) Calculate the acceleration: Step 2: Use the kinematic equation v = u + at. 0 m/s = 20 m/s + a(10 s) -20 m/s = 10a s a = -20 m/s10 s a = -2 m/s^2 The acceleration is -2 m/s^2. (The negative sign indicates deceleration). (ii) Calculate the distance travelled before stopping: Step 3: Use the kinematic equation s = ut + (1)/(2)at^2. s = (20 m/s)(10 s) + (1)/(2)(-2 m/s^2)(10 s)^2 s = 200 m - (1 m/s^2)(100 s^2) s = 200 m - 100 m s = 100 m Alternatively, using s = (u+v)/(2)t: s = 20 m/s + 0 m/s2 × 10 s s = 20 m/s2 × 10 s s = 10 m/s × 10 s s = 100 m The distance travelled before stopping is 100 m. --- 4. A body accelerates uniformly from 5 m/s to 25 m/s over a distance of 100 m. Calculate: (i) its acceleration (ii) time taken Given: Initial velocity, u = 5 m/s Final velocity, v = 25 m/s Displacement, s = 100 m (i) Calculate its acceleration: Step 1: Use the kinematic equation v^2 = u^2 + 2as. (25 m/s)^2 = (5 m/s)^2 + 2a(100 m) 625 m^2/s^2 = 25 m^2/s^2 + 200a m 625 - 25 = 200a 600 = 200a a = (600)/(200) m/s^2 a = 3 m/s^2 The acceleration is 3 m/s^2. (ii) Calculate the time taken: Step 2: Use the kinematic equation v = u + at. 25 m/s = 5 m/s + (3 m/s^2)t 25 - 5 = 3t 20 = 3t t = (20)/(3) s t ≈ 6.67 s The time taken is 6.67 s. --- 5. A particle moves such that its displacement is given by: S = 4t + 2t^2 (m). Determine: (i) its initial velocity (ii) its acceleration, and state its nature (iii) its velocity at 5 seconds Given: Displacement S = 4t + 2t^2. Recall: Velocity v = (dS)/(dt) Acceleration a = (dv)/(dt) Step 1: Find the velocity function by differentiating S with respect to t. v = (d)/(dt)(4t + 2t^2) v = 4 + 4t m/s (i) Determine its initial velocity: Step 2: Initial velocity is the velocity at t=0. v(0) = 4 + 4(0) v(0) = 4 m/s The initial velocity is 4 m/s. (ii) Determine its acceleration, and state its nature: Step 3: Find the acceleration function by differentiating v with respect to t. a = (d)/(dt)(4 + 4t) a = 4 m/s^2 The acceleration is 4 m/s^2. Since the acceleration is a constant positive value, its nature is uniform acceleration. (iii) Determine its velocity at 5 seconds: Step 4: Substitute t=5 s into the velocity function. v(5) = 4 + 4(5) v(5) = 4 + 20 v(5) = 24 m/s The velocity at 5 seconds is 24 m/s. --- 6. A car travelling at 15 m/s accelerates uniformly at 1.2 m/s^2. How long will it take to reach 30 m/s? Given: Initial velocity, u = 15 m/s Acceleration, a = 1.2 m/s^2 Final velocity, v = 30 m/s Time, t = ? Step 1: Use the kinematic equation v = u + at. 30 m/s = 15 m/s + (1.2 m/s^2)t 30 - 15 = 1.2t 15 = 1.2t t = (15)/(1.2) s t = 12.5 s It will take 12.5 s. --- 7. A particle starts from rest and accelerates uniformly at 4 m/s^2 for 5 seconds then immediately decelerates at 2 m/s^2 until it comes to rest. Find the total distance travelled by the particle. This problem has two phases: acceleration and deceleration. Phase 1: Acceleration Given: Initial velocity, u_1 = 0 m/s (starts from rest) Acceleration, a_1 = 4 m/s^2 Time, t_1 = 5 s Step 1: Calculate the final velocity of Phase 1 (v_1). This will be the initial velocity for Phase 2. v_1 = u_1 + a_1t_1 v_1 = 0 m/s + (4 m/s^2)(5 s) v_1 = 20 m/s Step 2: Calculate the distance travelled in Phase 1 (s_1). s_1 = u_1t_1 + (1)/(2)a_1t_1^2 s_1 = (0 m/s)(5 s) + (1)/(2)(4 m/s^2)(5 s)^2 s_1 = 0 + (1)/(2)(4 m/s^2)(25 s^2) s_1 = 2 m/s^2 × 25 s^2 s_1 = 50 m Phase 2: Deceleration Given: Initial velocity, u_2 = v_1 = 20 m/s Acceleration (deceleration), a_2 = -2 m/s^2 Final velocity, v_2 = 0 m/s (comes to rest) Step 3: Calculate the distance travelled in Phase 2 (s_2). Use the kinematic equation v_2^2 = u_2^2 + 2a_2s_2. (0 m/s)^2 = (20 m/s)^2 + 2(-2 m/s^2)s_2 0 = 400 m^2/s^2 - 4s_2 m/s^2 4s_2 m/s^2 = 4